java - GroupCount排序依据

下面的simplegraph-core测试套件中的Unit测试代码显示了机场的区域数量，但是没有按我期望的那样进行排序。

结果开始于:

NZ-BOP=  3
MZ-A=  1
MZ-B=  1
IN-TN=  5
MZ-N=  1
PW-004=  1
MZ-I=  2
BS-FP=  1
IN-TR=  1
MZ-T=  1
BJ-AQ=  1
GB-ENG= 27

我看着

Gremlin group by vertex property and get sum other properties in the same vertex

Gremlin query for groupcount of last value

并在标记为gremlin的问题中搜索“GroupCount”无济于事

需要什么来解决订购问题？

单元测试
另请参阅https://github.com/BITPlan/com.bitplan.simplegraph/blob/master/simplegraph-core/src/test/java/com/bitplan/simplegraph/core/TestTinkerPop3.java

  @Test
  public void testSortedGroupCount() throws Exception {
    Graph graph = getAirRoutes();
    GraphTraversalSource g = graph.traversal();
    Map<Object, Long> counts = g.V().hasLabel("airport").groupCount()
        .by("region").order().by(Order.decr).next();
    assertEquals(1473, counts.size());
    for (Object key : counts.keySet()) {
      System.out.println(String.format("%s=%3d", key, counts.get(key)));
    }
  }

最佳答案

您需要使用values作用域来订购local:

g.V().hasLabel("airport").
  groupCount().
    by("region").
  order(local).
    by(values, Order.decr)

在local范围内，您可以在当前遍历器中进行排序(即在遍历中对每个Map的内容进行排序)。

  @Test
  public void testSortedGroupCount() throws Exception {
    Graph graph = getAirRoutes();
    GraphTraversalSource g = graph.traversal();
    Map<Object, Long> counts = g.V().hasLabel("airport").groupCount()
        .by("region").order(Scope.local).by(Column.values,Order.decr).next();
    // https://stackoverflow.com/a/49361250/1497139
    assertEquals(1473, counts.size());
    assertEquals("LinkedHashMap",counts.getClass().getSimpleName());
    debug=true;
    if (debug)
      for (Object key : counts.keySet()) {
        System.out.println(String.format("%s=%3d", key, counts.get(key)));
      }

  }

然后将显示:

US-AK=149
AU-QLD= 50
CA-ON= 46
CA-QC= 44
PF-U-A= 30
US-CA= 29

SimpleGraph

java - GroupCount排序依据